update 2.2-2 2.2-3 2.2-4

Author: mcoderh

C02-Getting-Started/2.2.md

@@ -14,6 +14,10 @@ Consider sorting n numbers stored in array A by first finding the smallest eleme
 ### `Answer`
 ![pic](./repo/s2/1.png)
 
+Loop invariant: at the start of each iteration of the outer for loop, the subarray A[1..j-1] consists of the j-1 smallest elements in the array A[1..n] and this subarray is in sorted order.
+
+After the first n-1 elements, the subarray A[1..n-1] contains the smallest n-1 elements, sorted, and therefore element A[n] must be the largest element
+
 时间都是Θ(![](http://latex.codecogs.com/gif.latex?n^2))
 
 
@@ -22,10 +26,10 @@ Consider sorting n numbers stored in array A by first finding the smallest eleme
 Consider linear search again (see Exercise 2.1-3). How many elements of the input sequence need to be checked on the average, assuming that the element being searched for is equally likely to be any element in the array? How about in the worst case? What are the average-case and worst-case running times of linear search in Θ-notation? Justify your answers.
 
 ### `Answer`
-* 平均情况应该要查找n/2个元素
+* 平均情况应该要查找(n+1)/2个元素
 * 最坏情况是n个
 
-* Assuming equal probability of occurrence, average number of elements which need to be checked is n/2. Running time is Θ(n)
+* Assuming equal probability of occurrence 1/n, average number of elements which need to be checked is 1/n * (1 + 2 + ... +n) = (n+1)/2. Running time is Θ(n)
 * Worst case, the element to search is dead last in the array. In that case n elements need to be searched. Running time is Θ(n)
 
 所以都是Θ(n)
@@ -36,9 +40,9 @@ Consider linear search again (see Exercise 2.1-3). How many elements of the inpu
 How can we modify almost any algorithm to have a good best-case running time?
 
 ### `Answer`
-投机取巧，往最好情况去想
+算法开始检测输入数据，若符合特殊条件则输出事先计算好的结果
 
-One could check whether the problem is already solved (when sorting, for instance, check if input is already sorted).
+Modify the algorithm so it checks whether the input satisfies some special case condition. If it does, output a pre-computed answer.